99.4 thou if my maths is correct.

If the slot width is 2W and the ball diameter = 2R and the angle of the slot sides to the vertical = A:

The ball contacts the sides a height of R.sinA below the ball centre.

The distance across the slot at the contact points is 2R.cosA, so the height below the top of the slot is

(W-R.cosA) / tanA, giving the height of the top of the ball above the top of the slot as

H = R + R.sinA – (W-R.cosA) / tanA

For R = 0.5, W = 0.625 and A = 90-75 = 15, that gives the answer above.

Shoot me down if I'm wrong.

Rob

My answer is clearly wrong, because half of 1.125 is not 0.625

Edited By RobCox on 16/12/2021 12:20:56

Jason, I agree with your method, but I get the height of the contact below the top (x in your diagram) to be 0.0795/tan15 = 0.2968, which gives me a revised answer of 0.3326.

Thank goodness I haven't got to worry about maths exams any more!

Nothing other than a cursory glance, but I think it needs a couple of simultaneous equations to sort out the contact point. Might look at it later…

I can't help but feel inferior to the brains knocking around on this forum. I'd almost certainly have had to google the technique; I am for some reason unable to retain mathematics techniques and such.

Well done to the examiner and to those who attempted an answer!

I cheated and solved it by construction!

The answer there is 9.625 feet

Brian

Ladder box wall problem…

It can be done, I spent many a happy hour trying to solve this one some years ago, and although I didn't find a solution myself the problem can be solved, here is one solution.

**LINK**

I have just looked at the link. It is the same two solutions turned through ninety degrees.

It would be interesting see if complex numbers would make the solving easier. Too old for that now.

JA

Surely the first problem has a simple approach.
A line drawn vertically up from the ball/wall contact point meets a line extended from the block surface to give an 'inverted' triangle. Base of that is 1/2 (width of VGroove minus DiameterBall). Tan(15deg) then gives the inverted height to subtract from ball radius?

The second problem I’m struggling with. I think the solution rests somewhere in the fact that the hypotenuse of the triangle above the cube is also X such that there are simultaneous equations with Cos(Theta)=x/10=(x-2)/x. Also Tan(Theta)=(x-2)/2 and Sin(Theta)=2/X. I guess one needs to be familiar with the Trig reciprocals, which gets beyond the math I learned.?

pgk

I could try and pretend I'm really whizzy with maths by giving a solution, but I'm not, so I'll instead refer you to a youtube channel I've watched from time to time. The channel creator's name is Presh Talwalker and he solves this exact problem in his video " the ladder and box problem – a classic problem".

Me, I just got bogged down in quadratic equations that looked like they might turn into quartics…

I solved both problems in a matter of minutes. I was going to show you all

I wrote the solutions down on a piece of paper

But the dog ate it or I left it on the bus ….sorry sir

Happy Xmas all

Derek

My solution to the ladder problem:

Enough said.

JA